LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Static relativistic finite gas normalization and hydrostatics

Statement

Let $I_{x^{0}} \subseteq R$ be an open interval and $B \subset R^{3}$ an open bounded coordinate box with compact closure and positive volume. On a neighborhood of $I_{x^{0}} \times \overline{B}$ , with $x^{0} = c t$ assigned metres and $t$ in seconds, prescribe the smooth static metric $g = - N (x)^{2} (d x^{0})^{2} + h_{i j} (x) d x^{i} d x^{j},$ where $h_{i j}$ is positive definite with eigenvalues bounded above and away from zero, and the lapse satisfies $0 < N_{0} \leq N (x) \leq N_{1} < \infty$ . The static unit observer is $u = N^{- 1} \partial_{x^{0}}$ , so $g (u, u) = - 1$ ; this is a prescribed background, not a solution sourced by the gas.

Use SI units: particle mass $m > 0$ is in $k g$ , $c > 0$ is in $m s^{- 1}$ , orthonormal momentum $\hat{p}$ is in $k g m s^{- 1}$ , and local energy $E$ is in $J$ . In any local orthonormal coframe, $p_{i} = e_{i}^{a} {\hat{p}}_{a}$ and $h_{i j} = \sum_{a} e_{i}^{a} e_{j}^{a}$ , so $d V_{h} d^{3} \hat{p} = d^{3} x d^{3} p_{i}$ ; orthonormal-frame changes are rotations and preserve $d^{3} \hat{p}$ . This defines a fibre measure without assuming a global frame. Use the fixed action reference $h_{*} > 0$ in $J s$ and $β_{K} > 0$ in $J^{- 1}$ : $E (\hat{p}) = \sqrt{m^{2} c^{4} + c^{2} ∣ \hat{p} ∣^{2}}, H_{K} (x, \hat{p}) = N (x) E (\hat{p}), Z_{1} (β_{K}) = \int_{B} \int_{R^{3}} e^{- β_{K} H_{K}} \frac{d V_{h} d^{3} \hat{p}}{h_{*}^{3}} .$ Then $0 < Z_{1} < \infty$ , with all polynomial momentum moments and compact-positive- $β_{K}$ derivatives finite. For each fixed particle count $M \geq 1$ , the classical ideal-gas preparation with Gibbs factor $1 / M!$ has $Z_{M} = Z_{1}^{M} / M!$ ; a grand fugacity preparation with dimensionless $z > 0$ has $Ξ = \exp (z Z_{1})$ .

Define $f (x, \hat{p}) = z h_{*}^{- 3} e^{- β_{K} H_{K}}$ and the proper local densities $n (x) = \int f d^{3} \hat{p}, ε (x) = \int E f d^{3} \hat{p}, P (x) = \frac{1}{3} \int \hat{p} \cdot \nabla_{\hat{p}} E f d^{3} \hat{p} .$ Their units are $m^{- 3}$ , $J m^{- 3}$ , and $J m^{- 3}$ , respectively. With $β_{l o c} = β_{K} N (x)$ and $T_{K} = (k_{B} β_{K})^{- 1}$ , the exact model identities are $T_{l o c} (x) = \frac{1}{k_{B} β_{l o c}} = \frac{T_{K}}{N (x)}, P = n k_{B} T_{l o c}, \partial_{i} P = - (ε + P) \partial_{i} \log N .$ If $z = e^{β_{K} μ_{K}}$ with $μ_{K}$ in joules, the local chemical parameter is $μ_{l o c} = μ_{K} / N$ . No Einstein equation, self-consistent gas backreaction, interacting quantum trace, or arbitrary curved-space mode spectrum is asserted.

Facts & Assumptions

Given: A bounded positive-volume region, smooth prescribed static metric data satisfying the stated uniform bounds, $m, c, h_{*}, k_{B}, β_{K}, z > 0$ , and the classical ideal-gas preparation.

[F1]

The phase-reference item fixes the action unit $h_{*}$ and the convention that a phase integral is dimensionless after division by its corresponding power. (Finite gravitational phase and entropy reference)

[F2]

If an integrable section is differentiable in a parameter and its derivative has an integrable uniform majorant on a neighborhood, differentiation passes through the integral. (Differentiation under the integral sign)

[F3]

Tonelli factors nonnegative measurable integrals on products of sigma-finite measure spaces. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1F1algebra

In a local orthonormal coframe, $p_{i} = e_{i}^{a} {\hat{p}}_{a}$ and $h_{i j} = \sum_{a} e_{i}^{a} e_{j}^{a}$ , so $∣ \det (e_{i}^{a}) ∣ = \sqrt{\det h}$ . Hence $d^{3} p_{i} = \sqrt{\det h} d^{3} \hat{p}$ and $d^{3} x d^{3} p_{i} = d V_{h} d^{3} \hat{p}$ . The transition matrix between orthonormal coframes is orthogonal and has absolute determinant one, so this fibre measure is independent of the chosen local coframe.

1.2F1F2F3givenalgebra

In the orthonormal momentum variables, $E (\hat{p}) \geq c ∣ \hat{p} ∣$ , so $e^{- β_{K} N E} \leq e^{- β_{K} N_{0} c ∣ \hat{p} ∣}$ . The right side is integrable on $R^{3}$ : $∣ \hat{p} ∣ \geq (∣ {\hat{p}}_{1} ∣ + ∣ {\hat{p}}_{2} ∣ + ∣ {\hat{p}}_{3} ∣) / \sqrt{3}$ reduces its integral to a product of three finite one-dimensional exponential integrals. The same comparison with polynomial factors gives every energy moment and a common majorant for $β_{K}$ in compact positive intervals. Each $β_{K}$ derivative of the integrand is $(- H_{K})^{j} e^{- β_{K} H_{K}}$ ; [F2] therefore gives $Z_{1}^{(j)} = \int (- H_{K})^{j} e^{- β_{K} H_{K}} d V_{h} d^{3} \hat{p} / h_{*}^{3}$ for every finite order. The proper volume of $\overline{B}$ is finite by compactness and the metric eigenvalue bounds, and positive because $B$ has positive volume and $\det h > 0$ . Thus $Z_{1}$ is finite and strictly positive. For $M$ independent labelled particles, Tonelli [F3] factors the product integral; the fixed classical indistinguishable-particle counting convention divides by $M!$ , giving $Z_{M} = Z_{1}^{M} / M!$ . Summing grand weights yields $\sum_{M \geq 0} (z Z_{1})^{M} / M! = e^{z Z_{1}}$ .

2.1givenstep 1.2algebra

At a fixed $x$ , put $β_{l o c} = β_{K} N (x)$ . On a momentum cube $[- R, R]^{3}$ , integrate $\sum_{i = 1}^{3} \partial_{{\hat{p}}_{i}} ({\hat{p}}_{i} e^{- β_{l o c} E}) = 3 e^{- β_{l o c} E} - β_{l o c} (\hat{p} \cdot \nabla_{\hat{p}} E) e^{- β_{l o c} E} .$ The constant factor $z h_{*}^{- 3}$ in $f$ cancels from the moments. Here $\nabla_{\hat{p}} E = c^{2} \hat{p} / E$ and $\hat{p} \cdot \nabla_{\hat{p}} E \leq c ∣ \hat{p} ∣$ , so the divergence terms are integrable under the exponential bound of step 1.2. On a face with $∣ {\hat{p}}_{i} ∣ = R$ , $∣ \hat{p} ∣ \geq R$ and $∣ \hat{p} ∣ \geq (∣ {\hat{p}}_{j} ∣ + ∣ {\hat{p}}_{k} ∣) / \sqrt{2}$ , hence $∣ \hat{p} ∣ \geq R / 2 + (∣ {\hat{p}}_{j} ∣ + ∣ {\hat{p}}_{k} ∣) / (2 \sqrt{2})$ . Each face integral is therefore bounded by a constant times $R e^{- β_{l o c} c R / 2} \int_{R^{2}} e^{- β_{l o c} c (∣ y ∣ + ∣ z ∣) / (2 \sqrt{2})} d y d z$ , which tends to zero. The integral of the divergence over $R^{3}$ is zero, and division by $3 β_{l o c}$ gives $P = n / β_{l o c} = n k_{B} T_{l o c}$ .

3.1F1F2step 1.2step 2.1algebra∎

For each spatial coordinate $x^{i}$ , $\partial_{i} f = - β_{K} (\partial_{i} N) E f$ . The derivatives of the lapse are bounded on the compact closure, $N \geq N_{0}$ , and $E f$ has the integrable majorant from step 1.2. Hence [F2] gives $\partial_{i} n = - β_{K} (\partial_{i} N) ε$ . Differentiating $P = n / (β_{K} N)$ and substituting this identity yields $\partial_{i} P = - \frac{ε}{N} \partial_{i} N - \frac{n}{β_{K} N^{2}} \partial_{i} N = - (ε + P) \partial_{i} \log N .$ Finally, rewriting $β_{K} (N E - μ_{K})$ as $β_{l o c} (E - μ_{l o c})$ gives $μ_{l o c} = μ_{K} / N$ .

Static relativistic finite gas normalization and hydrostatics

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources